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题目:
给你两个非负整数
low
和high
。请你返回low
和high
之间(包括二者)奇数的数目。示例 1:
输入:low = 3, high = 7 输出:3 解释:3 到 7 之间奇数数字为 [3,5,7] 。示例 2:
输入:low = 8, high = 10 输出:1 解释:8 到 10 之间奇数数字为 [9] 。提示:
0 <= low <= high <= 10^9
第一种方法,直接判断奇数偶数,是奇数,计数器++;
- class Solution {
- public int countOdds(int low, int high) {
-
- int cnt = 0;
-
- if(low%2!=0){
- while(low<=high){
- cnt++;
- low+=2;
- }
- }
- if(low%2==0){
- low++;
- while(low<=high){
- cnt++;
- low+=2;
- }
- }
-
- return cnt;
- }
- }
第二种,列出所有情况,high-low=0;low奇,high偶;high奇,low偶;low、high全奇全偶。
- class Solution {
- public int countOdds(int low, int high) {
- int cnt = 0;
- if(high-low==0){
- if(low%2==0)return cnt;
- return cnt+1;
- }
- if(low%2==0&&high%2==0){
- cnt+=(high-low)/2;
- }else if(low%2!=0&&high%2!=0){
- cnt+=((high-low)/2+1);
- }else{
- cnt+=(high-low+1)/2;
- }
- return cnt;
- }
- }
第三种,一行代码秒杀!
- class Solution {
- public int countOdds(int low, int high) {
- return (high+1)/2 - low/2;
- }
- }
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